ASAP PLSSS!!!! A container holds 13.8 grams of lithium metal. How many moles is this?

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w

Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃ ΔH = –183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂ ΔH = 183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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7 0

2 years ago

For the balanced chemical reaction

musickatia [10]

Answer:

150

Explanation:

C₄H₂OH + 6O2 → 4CO2 + 5H₂O

We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:

100 molecules CO₂ * $\frac{6moleculesO_2}{4moleculesCO_2}$ = 150 molecules O₂

150 molecules of O₂ would produce 100 molecules of CO₂.

5 0

3 years ago

Select all of the following that are recessive alleles.

const2013 [10]

All of the lower case letters are recessive

8 0

2 years ago

If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the

Alina [70]

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

3 0

3 years ago